For the spontaneous process `2I(g)rarr I(2)(g)`, the sign of `DeltaH` and `DeltaS` respectively are

Options

A. `+ve,-ve`
B. `+ve,+ve`
C. `-ve,-ve`
D. `-ve,+ve`

Answer

To determine the signs of ΔH (change in enthalpy) and ΔS (change in entropy) for the spontaneous process \( 2I(g) \rightarrow I_2(g) \), we will analyze the reaction step by step.

### Step 1: Analyze the Reaction
The reaction involves the conversion of two moles of iodine gas (I) into one mole of diatomic iodine gas (I2).

### Step 2: Determine the Sign of ΔH
- **Enthalpy Change (ΔH)**:
- This process is exothermic, meaning that it releases heat. When a reaction releases heat, the change in enthalpy (ΔH) is negative.
- Therefore, for the reaction \( 2I(g) \rightarrow I_2(g) \), we conclude that:
\[ \Delta H < 0 \quad (\text{negative}) \]

### Step 3: Determine the Sign of ΔS
- **Entropy Change (ΔS)**:
- Entropy is a measure of the disorder or randomness of a system. In this reaction, we start with 2 moles of gaseous iodine (I) and end with 1 mole of diatomic iodine (I2).
- Since the number of gas molecules decreases from 2 to 1, the randomness or disorder of the system decreases.
- Therefore, the change in entropy (ΔS) is negative:
\[ \Delta S < 0 \quad (\text{negative}) \]

### Conclusion
For the spontaneous process \( 2I(g) \rightarrow I_2(g) \):
- The sign of ΔH is negative.
- The sign of ΔS is negative.

Thus, the final answer is:
\[ \Delta H < 0 \quad \text{and} \quad \Delta S < 0 \]